# Simple Substituting and Rearranging

1. Apr 18, 2013

### Caccioppoli

Hello,
may someone be so kind to explain how to arrive, step by step, from equation 23 to 28?
I simply cannot arrive at the result, no matter what, and I cannot explain how they got δ.

See the attached image.

Thank you very much.
Mario

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2. Apr 18, 2013

### Simon Bridge

All the steps appear to be explained by reference to a previous relation (not pictured) except for the last one ... where an approximation has been used.

i.e. eq23-24 requires eq17.

3. Apr 19, 2013

### Caccioppoli

I would like to understand only the approximation.
Maybe I could have got the logic behind: "approximating a cubic polynomial with a quadratic"; but if I substitute eq26 in 25 I get a different result (e.g. delta^3 terms).
eq24 may be taken as it is, take phi as just

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4. Apr 19, 2013

### Simon Bridge

OK - to avoid having to squint at the attachment - and so others can have easier input:
$$DC_sAt = A\left ( C_0-\frac{C_s}{2} \right )\frac{x^{\prime 2}}{2} -\frac{c}{6}\left ( C_0-\frac{7}{6}C_s \right )x^{\prime 3} \qquad \text{...(23)}\\ \text{...from eqn(17), \phi as a function of x^\prime is obtained as follows:}\\ \phi = \frac{ADC}{x^\prime + \frac{cx^{\prime 3}}{2A}}=\frac{ADC_s}{x^\prime}-\frac{cDC_s}{2} \qquad \text{...(24)}\\ \text{...from eqn(23), 1/x^\prime as a function of t is obtained:}\\ \text{Put x^\prime=1/q into eqn(27) then:}\\ \frac{2DC_s tq^3}{C_0-\frac{C_s}{2}}=q-\frac{c}{3A}\frac{C_0-\frac{7}{6}C_s}{C_0-\frac{C_s}{2}}\qquad \text{...(25)}\\ \text{Since one is only interested in approximate solutions for a short time,}\\ \text{ the following equation:}\\ q=\left ( \frac{2DC_st}{C_0-\frac{C_s}{2}}\right )^{-1/2}+\delta \qquad \text{...(26)}\\ \text{... may be substituted into eqn(25) to give:}\\ \delta = -\frac{c}{6A}\frac{C_0-\frac{7}{6}C_s}{C_0-\frac{C_s}{2}}\qquad \text{...(27)}\\ \text{if eqn(27) is substituted into eqn(26), and eqn(26) is further substituted into eqn(24),}\\ \text{ one obtains:}\\ \phi = A\left [ DC_s\left ( C_0-\frac{C_s}{2} \right )\frac{1}{2t} \right ]^{1/2} -\frac{4}{9}cDC_s\frac{3C_0-2C_s}{2C_0-C_s}\qquad \text{...(28)}$$
... so it is eqn 26 that is tricky?
The idea is to approximate for small time frames.

The form of the equation 25 is $\alpha q^3 = q - \beta$ ... which is a cubic equation.
Solve it for q, without making an approximation and compare.
They appear to have approximated it as $(q-\delta)^2 = \alpha$ and that delta is a parameter to be fitted.

Note: between eqn (24) and (25), there is a reference to eqn(27) which I suspect is a typo.

5. Apr 19, 2013

### Caccioppoli

Ahahaha for the "squint" , and thank you for giving others a readable input (sorry if I had not uploaded a bigger image).

By the way:

1) I can't figure why delta is beta/3 (using your notation for the terms) and I have no clue on how to fit delta

2) Also if I use that expressions in eq26 and eq27 for q and delta, putting them into phi (eq24) do not give 28: only the first term is correct (the one with the square root) and there is also a cDCs/2 which is not multiplicated to any C0; is there another approximation?

PS
That typo is actually a reference to eq24.

PS2

Last edited: Apr 19, 2013
6. Apr 19, 2013

### Simon Bridge

NO worries - typing it out makes sure I've read it properly ... putting it in latex means that others can copy and paste from what I wrote (just hit "quote" or "multi-quote" under the post to get the markup).

... it is a fitted term.
The usually default approximation is a Taylor series - just one or two orders.

7. Apr 20, 2013

### Caccioppoli

It was my mistake in the substitution of q+delta inside eq24, I've found that using eq26 and eq27 do actually bring to eq28, so no problem between eq26 and 28.

Last thing I would like to know is how that fitted term came out... I knew that every approximation should be based on Taylor series (!!!) any clue?

Basically, the tricky thing is eq26.

8. Apr 20, 2013

### Simon Bridge

I didn't actually do it ... my aim was to guide you to the realization.
Worked pretty well this time ;)

Basically - since it is an approximation, don't expect the substitutions to match exactly.
If you compare the exact approach (and do so by masking the distracting variables like I did) with what they got, you should be able to see where it's coming from.

Note: the other common approximation is the binomial one.
That's actually the first one I thought of since t has to be very small.
Also, sometimes, people use a power-series.

It's very naughty of them not to say which approximation method is being used... but they probably figure it would be obvious so look to one of the standards or methods the same author has used before.

9. Apr 22, 2013

### Caccioppoli

I haven't still figured out how to arrive at the result...

I have
$q=aq^3+b$ [eq#1]

that can somehow (?) be approximated with $q=a^{-0.5} + \delta$ [eq#2]

with $\delta=b/3$

Why is $\delta=b/3$?

10. Apr 22, 2013

### Simon Bridge

you have $aq^3-q+b=0$
and you want an approximation, it says, for small times (remember that a is a function of time).

does that mean that aq^3 is small?
q=1/x' ... how does x' depend on time, in general?

The approximation they came up with is $(q-\delta)^2 = a$
So they are saying that the cubic can behave like a quadratic for a short time.

You approaches towards understanding this is either to
(a) complete the calculation without the approximation and see what happens
(b) try to find the quadratic that behaves like the cubic for the short time in question
(c) try the other three approximation methods to see how they apply

The key appears to be figuring out what they mean by "a short time".
Very often in these books the reason why a particular approximation was chosen is unclear - the author leaves out all the trial and error involved in picking a useful one.

11. Apr 23, 2013

### Caccioppoli

You are right in stating there is a relation between x' and time.
The only thing I can tell reading the article is that delta x' are small if I consider small times (still not having an explicit relationship between x' and t)... but how could this relate with (x')^-3 or q^3 ?

Anyway, q^3 is undoubtedly small.

12. Apr 23, 2013

### Simon Bridge

If x' is small for small times, the q is large.
What about $tq^3$?

In general, if x is very small, then x^2 is even smaller and may be ignored.
Some quantity being small is particularly important for the binomial approximation.

Also note that if you integrate a quadratic, you end up with a 1/3 factor out front.

A Taylor series about t=0 would need us to know what q(0) was.
BUt I can see what you mean - the approx used is far from obvious.

Last edited: Apr 23, 2013
13. Apr 23, 2013

### Caccioppoli

Simon Bridge, I have managed to reach the solution, THANK YOU SO MUCH!

Premise: Having a look at eq 27 and using your notation we have that

δ is -b/2 not b/3 (sorry for the mistake )

So we have two equations, basically:

$aq^3-q+b=0$ and $q=a^{-0.5}+\delta$ with a small δ

Writing those two equations in a system, we have that the δ that satisfy both equations is a δ that satisfy this 3rd order equation:

$a\delta^3+3a^{-0.5}\delta^2+2\delta+b=0$

ignoring third order and second order terms in δ we have simply

$2\delta+b=0$ so the result $\delta=-b/2$

Last edited: Apr 23, 2013
14. Apr 23, 2013

### Caccioppoli

There is a clearer way to reach the result

eq.#1 is $aq^3+b=q$

eq.#1 would be simpler if b=0, the zero-order approximation is

$q_0=aq_0^3$ so $q_0=a^{-0.5}$

The next order (1st order) approximation is

$q=q_0 + \delta$

It is assumed that $\delta$ is small in comparison to $q_0$ so that all the terms in $\delta^2$ and $\delta^3$ are discarded and

$q^3=(q_0+\delta)^3≈q_0^3+3q_0^2\delta$

So that eq.#1 becomes

$q_0+\delta=a[q_0^3+3q_0^2\delta]+b$

Recalling the zero-order approximation we have that

$\delta=a[3\delta q_0^2]+b$

then

$\delta=a3\delta a^{-1}+b=3\delta+b$

The solution is $\delta=-b/2$

15. Apr 23, 2013

### Simon Bridge

Good on yer mate.
It was all you - I just pointed in likely looking directions.

I was beginning to worry I'd have to actually do it myself.
When you get a lot more experience, you will be expected to come up with these kinds of approximations yourself :)

Oh look - you managed to work out the reasoning for the particular approximation - double "well done" :D